What is the extreme f(x)=-2(4x-5)e^{4x} ?

The extreme f(x)=-2(4x-5)e^{4x} is Maximum(1,2e^4)

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extreme f(x)=-2(4x-5)e^{4x}

extreme points

f (x) = - 2 (4 x - 5) e^{4 x}

Maximum (1, 2 e^{4})

Solution steps

f^{'} (x) = - 2 (16 e^{4 x} x - 16 e^{4 x})

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 1

into

- 2 (4 x - 5) e^{4 x} : 2 e^{4}

Maximum (1, 2 e^{4})

f (x) = (2 cos (θ) + cos^{2} (θ))

f (x, y) = 2 x^{2} - x^{4} + 3 y^{2} - y^{3}

f (x) = 2 sin (x) cos (x)

f (x, y) = x^{2} + y^{2} - x y - x - y

f (x) = x^{2} + x y + y^{2} - 7 y + 16

What is the extreme f(x)=-2(4x-5)e^{4x} ?
The extreme f(x)=-2(4x-5)e^{4x} is Maximum(1,2e^4)